The reaction `2 N _(2) O _(5) (g) to NO _(2) (g) + O _(2) (g)` takes place in a closed contaner. It is found that the concentration of `NO _(2)` increases by `1. 6 xx 10 ^(-2) mol L ^(-1)` in four seconds . Calculate the rate of change of concentration of `N _(2) O _(5).`

Explanation

Reaction is ,
`2N_2O_5(g) rarr 4NO_2(g) +O_2(g)`
Rate=`-1/2(d[N_2O_5])/(dt)`
`=1/4(d[NO_2])/(dt)=(d[O_2])/(dt)`
`(d)/(dt)[NO_2]=(1.6xx10^(-2)" mol L"^(-1))/(4)`
`=4xx10^(-3)" mol L"^(-1)s^(-1)`
`(d)/(dt)[N_2O_5]=-2/4xx(d)/(dt)[NO_2]`
`=1/2(d)/(dt)[NO_2]`
`=-1/2xx4=10^(-3)" mol L"^(-1)s^(-1)`
`=-2xx10^(-3)" mol L"^(-1)s^(-1)`
Thus concentration of `N_2O_5`, decrease by `2xx10^(-3)" mol L"^(-1)s^(-1)`